How to evaluate $\int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1 + x}{1 - x}} \log\left(2x^2 - 2x + 1\right) \, dx$

Question

Question:

How to evaluate $$ \int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1 + x}{1 - x}} \log\left(2x^2 - 2x + 1\right) \, dx \approx -3.029597965685223 $$

Attempt:

$$ I \;=\;\int_{-1}^{1}\frac{1}{x}\,\sqrt{\frac{1+x}{1-x}}\;\ln\bigl(2x^2-2x+1\bigr)\,dx $$

Change of variable Set

$$ x = 1 - 2t,\qquad t\in[0,1], $$

so that $dx=-2\,dt$ and

$$ \frac{1+x}{1-x} = \frac{2-2t}{2t} = \frac{1-t}{t}, \quad 2x^2-2x+1 = 8t^2-4t+1. $$

One finds

$$ I = 2\int_{0}^{1}\frac{\,t^{-1/2}(1-t)^{1/2}\,}{1-2t} \,\ln\!\bigl(8t^2-4t+1\bigr)\,dt. $$ Next set $t=\sin^2\theta$, $\theta\in[0,\frac\pi2]$. Then

$$ dt = 2\sin\theta\cos\theta\,d\theta, \quad t^{-1/2}(1-t)^{1/2}\,dt = 2\cos^2\theta\,d\theta, \quad 1-2t = \cos2\theta. $$

Hence the integral becomes

$$ I =4\int_{0}^{\frac\pi2}\frac{\cos^2\theta}{\cos2\theta} \;\ln\!\bigl(2\cos^2(2\theta)-2\cos(2\theta)+1\bigr)\,d\theta. $$

Writing $\displaystyle \frac{\cos^2\theta}{\cos2\theta} =\frac12\Bigl(1+\frac1{\cos2\theta}\Bigr)$ splits $I=2J_1+2J_2$, where

$$ \begin{aligned} J_1 &= \int_{0}^{\frac\pi2} \ln\!\bigl(2\cos^2(2\theta)-2\cos2\theta+1\bigr)\,d\theta,\\ J_2 &= \int_{0}^{\frac\pi2} \frac{\ln\!\bigl(2\cos^2(2\theta)-2\cos2\theta+1\bigr)}{\cos2\theta}\,d\theta. \end{aligned} $$

I’m finding it difficult to progress any further. Is it possible to formulate any general type of generalization?

This integral is a variation of the famous integral discussed here:

Integral $\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right) \mathrm dx$

If we remove the numerator and move the denominator inside the logarithm, the integral appears.

Eight downvotes for no reason great.

The closed-form expression of the integral, given by Integreek, is  $$\pi \left(\coth^{-1} \sqrt{\phi} - 2 \cot^{-1} \sqrt{\phi} - \ln 2\right).$$

Edit 2: can someone explain why this question was closed again? Initially, other users said it lacked context, so I added the necessary context. But it’s still getting closed.

Answer 1

\begin{align}\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln(2x^2-2x+1)\mathrm dx&=\int_0^1\frac1x\left(\sqrt{\frac{1+x}{1-x}}\ln(2x^2-2x+1)-\sqrt{\frac{1-x}{1+x}}\ln(2x^2+2x+1)\right)\mathrm dx\\&=\underbrace{\int_0^1\frac{\ln(4x^4+1)}{\sqrt{1-x^2}}\mathrm dx}_{I_1}-\underbrace{\int_0^1\ln\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right)\frac{\mathrm dx}{x\sqrt{1-x^2}}}_{I_2}\end{align}

$I_2$ is actually half of the famous integral $\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right)\mathrm dx:$

\begin{align}\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right)\mathrm dx&=\int_0^1\ln\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right)\left(\sqrt{\frac{1+x}{1-x}}+\sqrt{\frac{1-x}{1+x}}\right)\frac{\mathrm dx}x\\&=2\int_0^1\ln\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right)\frac{\mathrm dx}{x\sqrt{1-x^2}}\\\implies I_2&=2\pi\cot^{-1}\sqrt\phi\end{align}

And $I_1$ can be found by using the result

$$\int_0^{\pi/2}\ln(\sin^2x+a)\mathrm dx=\pi(\sinh^{-1}\sqrt a-\ln2)$$

\begin{align}\frac{\mathrm d}{\mathrm da}\left(\int_0^{\pi/2}\ln(\sin^2x+a)\mathrm dx\right)&=\int_0^{\pi/2}\frac{\sec^2x~\mathrm dx}{(a+1)\tan^2x+a}\\&=\frac{\pi}{2\sqrt a\sqrt{a+1}}\\\implies \int_0^{\pi/2}\ln(\sin^2x+a)\mathrm dx&=\pi\int_0^a\frac{\mathrm d}{\mathrm dv}\left(\sqrt v\right)\frac1{\sqrt{\left(\sqrt v\right)^2+1}}\mathrm dv\\&=\pi\left(\sinh^{-1}\sqrt a-\ln2\right)\end{align}

$$I_1=\pi\left(-\ln2+\sinh^{-1}\left(\frac{1+i}2\right)+\sinh^{-1}\left(\frac{1-i}2\right)\right)$$

Using $\sinh^{-1}x+\sinh^{-1}y=\sinh^{-1}\left(x\sqrt{y^2+1}+y\sqrt{x^2+1}\right)$ and on simplifying using the identities $\phi^{\pm3}=\sqrt5\pm2,\phi^2=\frac{3+\sqrt5}2,\phi^2+1=\phi$, we have

$$I_1=\pi\left(\coth^{-1}\sqrt\phi-\ln2\right)$$

Hence, $$\boxed{\begin{align}\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln(2x^2-2x+1)\mathrm dx&=\pi\left(\coth^{-1}\sqrt\phi-2\cot^{-1}\sqrt\phi-\ln2\right)\end{align}}$$

Answer 2

I think that it could be "easier" to do $$\sqrt{\frac{1+x}{1-x}}=t \quad \implies \quad x=\frac{t^2-1}{t^2+1}\quad \implies \quad dx=\frac{4 t}{\left(t^2+1\right)^2}\,dt$$ which gives $$I=4\int_0^\infty \frac{t^2}{t^4-1} \log \left(\frac{t^4-2 t^2+5}{\left(t^2+1\right)^2}\right)\,dt$$

Then, decompose the logarithm using $$t^4-2t^2+5=(t^2-(1-2 i))\,(t^2-(1+2 i))$$ $$(t^2+1)^2=(t-i)^2\,(t+i)^2$$ as well as $$\frac{t^2}{t^4-1}=-\frac{i}{4 (t-i)}+\frac{i}{4 (t+i)}-\frac{1}{4 (t+1)}+\frac{1}{4 (t-1)}$$

The antiderivatives are not so difficult. The problm is that the result for the definite integral is a monster (a bunch of logarithms and polylogarithms of complex numbers).

Have look here for the antiderivative. If the link is broken, give Wolfram Alpha the following

Integrate[(4*t^2*Log[(5 - 2*t^2 + t^4)/(1 + t^2)^2])/ (-1 + t^4),t]

Answer 3

From the "King's Rule" for definite integration, $$\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$$ we can replace $\cos(2\theta)$ with $\cos(2(\pi/2-\theta))=-\cos(2\theta)$, $$J_1=\int_{0}^{\frac\pi2}\ln(2\cos^2(2\theta)-2\cos(2\theta)+1)d\theta=\int_{0}^{\frac\pi2}\ln(2\cos^2(2\theta)+2\cos(2\theta)+1)d\theta$$ and then add both forms of $J_1$, $$J_1+J_1=\int_{0}^{\frac\pi2}(\ln(2\cos^2(2\theta)-2\cos(2\theta)+1)+\ln(2\cos^2(2\theta)+2\cos(2\theta)+1))d\theta$$ $$2J_1=\int_{0}^{\frac\pi2}\ln((2\cos^2(2\theta)-2\cos(2\theta)+1)(2\cos^2(2\theta)+2\cos(2\theta)+1))d\theta$$ $$J_1=\frac{1}{2}\int_{0}^{\frac\pi2}\ln(4\cos^4(2\theta)+1)d\theta$$ As far as I know there isn't a simple closed form solution to this (there might be a complicated one).